1985 AMC 8 Problem 8 — Solution | LIVE by Po-Shen Loh

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1985 AMC 8 Problem 8

Attempt Problem 8 of the 1985 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

If $a = -2,$ the largest number in the set

$\left\{ -3a,\ 4a,\ \dfrac{24}{a},\ a^2,\ 1 \right\}$

is

$-3a$

$4a$

$\dfrac{24}{a}$

$a^2$

$1$

Answer: A

Concepts:substitution

Difficulty rating: 730

Solution:

Substituting $a = -2$ gives the values $-3a = 6,$ $4a = -8,$ $\dfrac{24}{a} = -12,$ $a^2 = 4,$ and $1.$

The largest of these is $6,$ which is $-3a.$

Thus, the correct answer is A .

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